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Simultaneous equations Simultaneous equations (Meyer, p. 1)
“Three sheafs of a good crop, two sheafs of a mediocre crop, and one sheaf of a bad crop are sold for 39 dou. Two sheafs of good, three mediocre, and one bad are sold for 34 dou; and one good, two mediocre, and three bad are sold for 26 dou. What is the price received for each sheaf of a good crop, each sheaf of a mediocre crop, and each sheaf of a bad crop?”
Chapter VIII, Chiu-chang Suan-shu (Nine chapters on Arithmetic). circa 200 B.C.
Solution:
x 1 : good crop; x 2 : mediocre crop; x 3 : bad crop
Simultaneous equations Matrix form
3 x 1 + 2x 2 + x 3 = 39, 2 x 1 + 3x 2 + x 3 = 34, x 1 + 2x 2 + 3x 3 = 26.
x (^1) x (^2) x (^3)
Introduction
Matrix Analysis and Computation
ECE 210a, CMPSC 211a, ME 210a, Math 206a, ChE 211a
Roy Smith
5117 Harold Frank Hall (805) 893– [email protected]
Roy Smith: ECE 210a: 1.
Diagonal factorizations Interpretation:
The solution equations are: Rˆ x = ˆb
(^) x =
(^) A x =
(^) b,
T 2 T 1 A x = T 2 T 1 b.
So Rˆ = T 2 T 1 A, or equivalently, A = T 1 − 1 T 2 − 1 Rˆ = Qˆ R.ˆ
In general we can always find a unitary Q and upper triangular R such that, A = QR, (Q ∗^ Q = QQ ∗^ = I) then Rx = Q ∗^ b is a triangular system.
Diagonal factorizations Solution:
Simultaneous equations Matrix form
3 x 1 + 2x 2 + x 3 = 39, 2 x 1 + 3x 2 + x 3 = 34, x 1 + 2x 2 + 3x 3 = 26
x (^1) x (^2) x (^3)
3 x 1 + 2x 2 + x 3 = 39 9 x 2 + x 3 = 44 4 x 2 + 8x 3 = 39
(3r 2 − 2 r 1 ) (3r 3 − r 1 )
x (^1) x (^2) x (^3)
3 x 1 + 2x 2 + x 3 = 39 5 x 2 + x 3 = 44 68 x 3 = 175 (5r 3 − 4 r 2 )
x (^1) x (^2) x (^3)
x (^1) x (^2) x (^3)
So x 1 = 9 14 , x 2 = 4 14 and x 3 = 2 34.
Roy Smith: ECE 210a: 1.
Planetary orbit determination Steps in Gauss’s approach
− Select three observations: 1st January, 21st January and 11th February. Each observation is two angles (Ceres with respect to the background stars) and a time. − Calculate a nominal orbit matching the observations. − Taylor series approximations to the nonlinear differential equations. − Iterative refinement. − Adjust linearized versions of the orbit to minimize the sum of squares of all 22 observations.
Problem formulation: H is the nonlinear orbital prediction model.
ymodel =
RA of obs. # dec. of obs. #1. .. RA. of obs. # dec. of obs. #
^ =^ H(x.t)^ where^ x^ = orbital elements.
Planetary orbit determination Gauss’s approach
The orbit of a planet is defined by five variables (orbital elements)
− Two angles that orient the plane of the orbit with respect to the ecliptic plane. − The size and eccentricity of the orbit. Equivalently the length of the major and minor semi-axes. − The tilt of the main axis of the orbit with respect to that of Earth’s orbit.
Ecliptic plane
Sun
Ceres
P Earth
P
E
C
− Knowing the location on the orbit at a particular time completely specifies the location in space.
Roy Smith: ECE 210a: 1.
Planetary orbit determination Linearize the problem:
The nominal orbit is specified by x 0. Close to this,
H(x) ≈ H(x 0 ) + H (^) J (x − x 0 ), where H (^) J = ∂y ∂modelx (Jacobian).
Define δx = x − x 0 and δy = yobs − H(x 0 ). Then
J (^) linear (δx) := (δy^ −^ H^ J^ δx)^
T (^) R − (^1) (δy − H (^) J δx) 2 ≈^ J(x^0 +^ δx). The linearized cost function is close to the true cost function. And it’s quadratic.
The gradient of J (^) linear (δx) is, ∂J (^) linear (δx) δx =^ −H^ JT R^ −^1 (δy^ −^ H^ J^ δx).
Setting this to zero is equivalent to solving: H (^) JT R −^1 H (^) J δx = H TJ R −^1 δy Note the form: Ax = b.
It the nonlinear case we would apply this iteratively.
Planetary orbit determination Problem formulation: yobs are the actual (noisy) observations.
= yobs − H(x, t) # is the residual.
The cost function is J(x) = #T^ R 2 − 1 #. R is a weighting matrix.
Objective:
x (^) optimal = min x J(x).
Nominal orbit
1 Jan.
21 Jan.
11 Feb.
Nominal orbit
LS fit candidate orbits
ε j
Nominal orbit determination Least squares orbit fit 6 equations in 6 unknown parameters 44 equations in 6 unknown parameters
Roy Smith: ECE 210a: 1.
Vibrational spectroscopy Diagonalizing transform
Find a unitary U (U T^ U = U U T^ = I) and diagonal Λ =
λ 1 0
... 0 λN
(^) such that,
K = U ΛU T^ , so d^
2 dt^2
q (^1) ... q (^) N
= U ΛU T
q (^1) ... q (^) N
Define new coordinates (normal coordinates),
Q 1
Q N
= U T
q (^1) ... q (^) N
Then,
d 2 dt^2
Q. 1
Q N
= U T^ U ΛU T
q. (^1) .. q (^) N
Q. 1
Q N
(^). N independent oscillators.
Vibrational spectroscopy Newton’s second law:
mi^ d^
(^2) x (^) i dt^2 =^ −^
j
k (^) ij x (^) j.
Key feature: k (^) ij = k (^) ji.
Introduce mass-weighted coordinates: q (^) i = √mi x (^) i. d 2 q (^) i dt^2 =^ −^
j
√^ k^ ij mi mj^ q^ j^ =^
j
K (^) ij q (^) j.
Matrix form:
d 2 dt^2
q. (^1) .. q (^) N
K. 11 · · · K 1 N
K N 1 · · · K N N
q. (^1) .. q (^) N
= K
q. (^1) .. q (^) N
This is a multi-modal oscillator with modal frequency given by the square root of the eigenvalues of K.
The “shapes” of the modes are given by the eigenvectors.
Roy Smith: ECE 210a: 1.
Vibrational spectroscopy Normal modes:
The eigenvalues of K =
k/mO −k/√mO mC 0 −k/√mO mC 2 k/mC −k/√mO mC 0 −k/√mO mC k/mO
(^) are:
Eigenvalues Eigenvectors Eigenvectors λi (normal coord.) (physical coord.)
0 1 /m
√mO √mc √mO
(^1) /m
no restoring force(translational mode)
k/mO 1 /√ 2
(^1) /√ 2 mO
independent of (carbon doesn’t move)^ mC
km/mO mC 1 /√ 2 m
√mC − √ 2 √mO mC
(^1) /√ 2 m
√m C /mO −√ 2 √mO /mC mC /mO
carbon oscillatingbetween oxygens
(m is the total mass of the molecule: m = mC + 2mO ).
Vibrational spectroscopy Carbon Dioxide Example (1 dimensional)
k m
1 2 3 x
O m C^ m^ O
O C O
x x x
V = k 2 (x 1 − x 2 ) 2 + k 2 (x 3 − x 2 ) 2
k (^) ij = ∂^
2 V
∂x (^) i x (^) j^ ,^ i, j^ = 1,^2 ,^3.
[k (^) ij ] =
k −k 0 −k 2 k −k 0 −k k
(^) , and K =
k/mO −k/√mO mC 0 −k/√mO mC 2 k/mC −k/√mO mC 0 −k/√mO mC k/mO
Roy Smith: ECE 210a: 1.
Markov processes Markov process:
The probability of the pea being under a particular shell depends only on the shell it was previously under.
If p (^) i is the probability that the pea is under shell i, then, in going from shuffle k to k + 1,
p(k + 1) =
p 1 (k + 1) p 2 (k + 1) p 3 (k + 1) p 4 (k + 1)
p 1 (k) p 2 (k) p 3 (k) p 4 (k)
=^ Ap(k).
Note that A (^) ij ≥ 0 and
∑^4
i=
A (^) ij = 1.
Question 1: Applying the above iteratively gives, p(N ) = A N^ p(0).
Question 2: The solution would be given by,
klim−→∞ A^ k but unfortunately this limit does not exist in this case.
What can be said about the long term probabilities?
Shell game (Meyer, example 7.10.8)
A “pea” is hidden under one of four “shells” which are randomly shuffled according to the probability rules:
The pea must be moved with each shuffle.
Questions
- If we know where the pea starts, can we say anything about the probability of where it will end up after N moves?
- Is there a limiting value to the probability and does it depend on the pea’s starting shell?
Roy Smith: ECE 210a: 1.
